Answer
{$\sqrt {11},-4$} and {$-\sqrt {11},-4$}
Work Step by Step
Given: $3x^2+4y=17$ and $2x^2+5y=2$
Multiply the above equations with the coefficients.
we get $15x^2+20y=85$ and $8x^2+20y=8$
Now,$15x^2+20y-8x^2-20y=85-8$
Then, we have $7x^2=77$
This gives: $x=\pm \sqrt {11}$
a) At $x= \sqrt {11}$, we have $2(\sqrt {11})^2+5y=2 \implies y=-4$
b) At $x= -\sqrt {11}$, we have $2(-\sqrt {11})^2+5y=2 \implies y=-4$
Thus, the solutions set are:
{$\sqrt {11},-4$} and {$-\sqrt {11},-4$}
